Board 2025 Class 12 Maths Q31 Solution: Sketch y = |x + 3| and find the area between x = −6 and x = 0

The question

Sketch the graph of $y = |x + 3|$ and find the area of the region enclosed by the curve, $x$ -axis, between $x = -6$ and $x = 0$ ⁠, using integration.

Exam answer

\begin{aligned} y &= |x + 3| \\ &= \begin{cases} -(x + 3), & x < -3 \\ x + 3, & x \ge -3 \end{cases} \end{aligned}

Why?

By the definition of modulus, $|t| = t$ when $t \ge 0$ and $|t| = -t$ when $t < 0$ ⁠.

Here $t = x + 3$ ⁠, and $x + 3 \ge 0$ exactly when $x \ge -3$ ⁠. So the formula for $y$ changes at $x = -3$ ⁠.

The graph is two half-lines meeting at $(-3, 0)$ ⁠, through $(-6, 3)$ and $(0, 3)$ (graph below).

Why?

Each piece is a straight line. Plot a few points: $x = -6$ gives $y = |-3| = 3$ ⁠; $x = -3$ gives $y = 0$ ⁠; $x = 0$ gives $y = 3$ ⁠. The corner of the V is where $x + 3 = 0$ ⁠.

The graph of

y = |x + 3|

⁠; the region between

x = -6

and

x = 0

is shaded.

\begin{aligned} \text{Area} &= \int_{-6}^{0}y\,dx \\ &= \int_{-6}^{-3}-(x + 3)\,dx \\ &\quad + \int_{-3}^{0}(x + 3)\,dx \end{aligned}

Why?

The curve lies on or above the $x$ -axis, so the area between $x = a$ and $x = b$ is $\int_a^b y\,dx$ ⁠.

The formula for $y$ changes at $x = -3$ ⁠, so split the integral there and use the right formula on each piece.

\begin{aligned} \text{Area} &= -\left[\frac{(x + 3)^2}{2}\right]_{-6}^{-3} \\ &\quad + \left[\frac{(x + 3)^2}{2}\right]_{-3}^{0} \\ &= -\left(0 - \frac92\right) + \left(\frac92 - 0\right) \\ &= \frac92 + \frac92 \end{aligned}

Why?

An antiderivative of $x + 3$ is $\frac{(x + 3)^2}{2}$ ⁠, since

\frac{d}{dx}\frac{(x + 3)^2}{2} = x + 3

At $x = -3$ it is $0$ ⁠. At $x = -6$ and at $x = 0$ it is $\frac{3^2}{2} = \frac92$ ⁠.

\begin{aligned} x = -3&:\ \frac{(-3 + 3)^2}{2} = \frac{0^2}{2} = 0 \\ x = -6&:\ \frac{(-6 + 3)^2}{2} = \frac{(-3)^2}{2} = \frac92 \\ x = 0&:\ \frac{(0 + 3)^2}{2} = \frac{3^2}{2} = \frac92 \end{aligned}

Each bracket is (value at the top limit) minus (value at the bottom limit):

\begin{aligned} -\left(0 - \frac92\right) &= -\left(-\frac92\right) = \frac92 \\ \frac92 - 0 &= \frac92 \\[4pt] \frac92 + \frac92 &= \frac{18}{2} = 9 \end{aligned}

Check with the figure: the region is two triangles, each with base $3$ and height $3$ ⁠, so each has area $\frac12\cdot3\cdot3 = \frac92$ ⁠.

$\therefore$ Required area $= 9$ sq. units.

Where this comes from

Chapter 8, Application of Integrals: Area under y = |x + 3| by integration.

Question type: A modulus inside the integral

How to spot it: click to see

Type 6. A modulus inside the integral: split at the corner

Spot it: bars inside the integral, like $|x + 3|$ or $x|x|$ ⁠.

Here: $|x + 3|$ has its corner at $x = -3$ ⁠, between $-6$ and 0. Split there: use $-(x + 3)$ from $-6$ to $-3$ and $x + 3$ from $-3$ to 0, giving $\frac92 + \frac92 = 9$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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Q31: Sketch y=∣x+3∣y = |x + 3|y=∣x+3∣ and find the area between x=−6x = -6x=−6 and x=0x = 0x=0

Q31: Sketch $y = |x + 3|$ and find the area between $x = -6$ and $x = 0$